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Algebra Problems… Set 21 By Herbert I. Gross and Richard A. Medeiros

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Presentation on theme: "Algebra Problems… Set 21 By Herbert I. Gross and Richard A. Medeiros"— Presentation transcript:

1 Algebra Problems… Set 21 By Herbert I. Gross and Richard A. Medeiros
next Algebra Problems… Set 21 By Herbert I. Gross and Richard A. Medeiros © Herbert I. Gross

2 Problem #1a Find the values of x and y for which… 6x + 5y = 15
next Problem #1a Find the values of x and y for which… 6x + 5y = 15 5x + 4y = 20 © Herbert I. Gross 2

3 Problem #1b Find the values of x and y for which… 0.06x + 0.05y = 0.15
next Problem #1b Find the values of x and y for which… 0.06x y = 0.15 0.05x y = 0.20 © Herbert I. Gross 3

4 Problem #1c Find the values of x and y for which… 1/5 x + 1/6 y = 1/2
next Problem #1c Find the values of x and y for which… 1/5 x + 1/6 y = 1/2 1/4 x + 1/5 y = 1 © Herbert I. Gross 4

5 Problem #1d The equation of the line L1 is 6x + 5y = 15,
next Problem #1d The equation of the line L1 is 6x + 5y = 15, and the equation of the line L2 is 5x + 4y = 20. At what point do these two lines intersect? © Herbert I. Gross 5

6 Problem #2a Find the values of x and y for which… 2x + 3y = 111
next Problem #2a Find the values of x and y for which… 2x + 3y = 111 5x + 4y = 204 © Herbert I. Gross 6

7 What is the cost of each pear ?
next Problem #2b Two (equally priced) apples and three (equally priced) pears cost $1.11 while five apples and four pears cost $2.04. What is the cost of each pear ? © Herbert I. Gross 7

8 Problem #3a Find the values of x and y for which… x + y = 50
next Problem #3a Find the values of x and y for which… x + y = 50 7.5x + 5y = 300 © Herbert I. Gross 8

9 next Problem #3b Coffee beans worth $7.50 per pound are blended with coffee beans worth $5 per pound to make a mixture of 50 pounds worth $6 per pound. How many pounds of each mixture are there in the blend? © Herbert I. Gross 9

10 Problem #4 Find the values of x, y, and z for which… x + 3y + 5z = 111
next Problem #4 Find the values of x, y, and z for which… x + 3y + 5z = 111 y = 3x + 2 z = 2y + 1 © Herbert I. Gross 10

11 Problem #5 Find the values of x, y, and z for which… x + y + z = 2
next Problem #5 Find the values of x, y, and z for which… x + y + z = 2 2x + 3y + 3z = 5 3x + 4y + 2z = 1 © Herbert I. Gross 11


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